3.27.92 \(\int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{-45 x^3+18 x^3 \log (3)} \, dx\)

Optimal. Leaf size=17 \[ e^{\frac {16}{9 x^2 \left (-\frac {5}{2}+\log (3)\right )}} \]

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Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 6706} \begin {gather*} e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64*E^(32/(-45*x^2 + 18*x^2*Log[3])))/(-45*x^3 + 18*x^3*Log[3]),x]

[Out]

E^(-32/(9*(5*x^2 - 2*x^2*Log[3])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int -\frac {64 e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3 (-45+18 \log (3))} \, dx\\ &=\frac {64 \int \frac {e^{\frac {32}{-45 x^2+18 x^2 \log (3)}}}{x^3} \, dx}{9 (5-\log (9))}\\ &=e^{-\frac {32}{9 \left (5 x^2-2 x^2 \log (3)\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.88 \begin {gather*} e^{\frac {32}{9 x^2 (-5+\log (9))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64*E^(32/(-45*x^2 + 18*x^2*Log[3])))/(-45*x^3 + 18*x^3*Log[3]),x]

[Out]

E^(32/(9*x^2*(-5 + Log[9])))

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fricas [A]  time = 0.96, size = 18, normalized size = 1.06 \begin {gather*} e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \relax (3) - 5 \, x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="fricas")

[Out]

e^(32/9/(2*x^2*log(3) - 5*x^2))

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giac [A]  time = 0.21, size = 18, normalized size = 1.06 \begin {gather*} e^{\left (\frac {32}{9 \, {\left (2 \, x^{2} \log \relax (3) - 5 \, x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="giac")

[Out]

e^(32/9/(2*x^2*log(3) - 5*x^2))

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maple [A]  time = 0.07, size = 15, normalized size = 0.88

method	result	size
gosper	\({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \relax (3)-5\right )}}\)	\(15\)
derivativedivides	\({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \relax (3)-5\right )}}\)	\(15\)
default	\({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \relax (3)-5\right )}}\)	\(15\)
risch	\({\mathrm e}^{\frac {32}{9 x^{2} \left (2 \ln \relax (3)-5\right )}}\)	\(15\)
norman	\({\mathrm e}^{\frac {32}{18 x^{2} \ln \relax (3)-45 x^{2}}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-64*exp(32/(18*x^2*ln(3)-45*x^2))/(18*x^3*ln(3)-45*x^3),x,method=_RETURNVERBOSE)

[Out]

exp(32/9/x^2/(2*ln(3)-5))

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maxima [A]  time = 0.49, size = 14, normalized size = 0.82 \begin {gather*} e^{\left (\frac {32}{9 \, x^{2} {\left (2 \, \log \relax (3) - 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(32/(18*x^2*log(3)-45*x^2))/(18*x^3*log(3)-45*x^3),x, algorithm="maxima")

[Out]

e^(32/9/(x^2*(2*log(3) - 5)))

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mupad [B]  time = 1.55, size = 14, normalized size = 0.82 \begin {gather*} {\mathrm {e}}^{\frac {32}{9\,x^2\,\left (2\,\ln \relax (3)-5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*exp(32/(18*x^2*log(3) - 45*x^2)))/(18*x^3*log(3) - 45*x^3),x)

[Out]

exp(32/(9*x^2*(2*log(3) - 5)))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.88 \begin {gather*} e^{\frac {32}{- 45 x^{2} + 18 x^{2} \log {\relax (3 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-64*exp(32/(18*x**2*ln(3)-45*x**2))/(18*x**3*ln(3)-45*x**3),x)

[Out]

exp(32/(-45*x**2 + 18*x**2*log(3)))

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